# Thread: [Algebra?(rationalize)] I've never learnt this

1. ## [Algebra?(rationalize)] I've never learnt this

I never been taught this and it's on the exam papers, how do I do this?

2. Originally Posted by Cthul

I never been taught this and it's on the exam papers, how do I do this?
You'll need to get it of the form

$\frac{2(x^2 - 2x - 3) + R}{x^2 - 2x - 3}$ where $R$ is some remainder...

$\frac{2x^2 + 2x + 1}{x^2 - 2x - 3} = \frac{2x^2 - 4x - 6 + 6x + 7}{x^2 - 2x - 3}$

$= \frac{2(x^2 - 2x - 3) + 6x + 7}{x^2 - 2x - 3}$

$= \frac{2(x^2 - 2x - 3)}{x^2 - 2x - 3} + \frac{6x + 7}{x^2 - 2x - 3}$

$= 2 + \frac{6x + 7}{x^2 - 2x - 3}$

$= 2 + \frac{6x + 7}{(x - 3)(x + 1)}$.

Now you need to break down the remainder using the method of partial fractions.

Let $\frac{A}{x - 3} + \frac{B}{x + 1} = \frac{6x + 7}{(x - 3)(x + 1)}$

$\frac{A(x + 1) + B(x - 3)}{(x - 3)(x + 1)} = \frac{6x + 7}{(x - 3)(x + 1)}$

$A(x + 1) + B(x - 3) = 6x + 7$

$Ax + A + Bx - 3B = 6x + 7$

$(A + B)x + A - 3B = 6x + 7$.

So

$A + B = 6$

$A - 3B = 7$.

Subtracting the second equation from the first gives

$4B = -1$

$B = -\frac{1}{4}$.

Since $A + B = 6$

$A - \frac{1}{4} = 6$

$A = \frac{25}{4}$.

So finally we have:

$\frac{2x^2 + 2x + 1}{x^2 - 2x - 3} = 2 + \frac{6x + 7}{(x - 3)(x + 1)}$

$= 2 + \frac{\frac{25}{4}}{(x - 3)} - \frac{\frac{1}{4}}{(x + 1)}$

$= 2 + \frac{25}{4(x - 3)} - \frac{1}{4(x + 1)}$.

3. Originally Posted by Cthul

I never been taught this and it's on the exam papers, how do I do this?
First, divide $x^2-2x-3$ into $2x^2+2x+1$

You will get a remainder

Hence $\frac{2x^2+2x+1}{x^2-2x-3}=2+\frac{remainder}{(x-3)(x+1)}$

since $x^2-2x-3=(x-3)(x+1)$

Now you need to express the remaining fraction as the sum of partial fractions

$\frac{remainder}{(x-3)(x+1)}=\frac{A}{x-3}+\frac{B}{x+1}$

Solve for A and B.

4. ## Thanks

This helped, thank you.
(Solved)